Trying to solve time of charge of supercapacitor with ode45

Question

Isaac Husny T 2020-4-29

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https://ww2.mathworks.cn/matlabcentral/answers/521549-trying-to-solve-time-of-charge-of-supercapacitor-with-ode45

评论： David Goodmanson 2020-5-1

fprintf(' Modelación del Supercapacitor');
%Inputs
R1=100;%Ohms
R2=100;%Ohms
RC=1000;%Ohms
C1=.00001;%Farads
C2=.0001;%Farads
DV=5;%Voltios
%Pasar a valores para sacar raices
A=(RC+R1)*R2*C1;
B=(RC+R1)*(C1/C2)+R2;
C=1;
%Eliminar raÌces imaginarias
   %fg=@(t,x)[x(2);(DV-B*x(2)-C*x(1))/A];
fg=@(t,x) [x(1);(DV-B*x(1)-C*x(2))/A];
tf = 5;
   %Polinomio
   %p=[A,B,C];
   %rp=roots(p);
   %rmin=min(abs(rp));
Dt=[0,tf];%De donde a donde queremos graficar
X0=[0,0];%Condiciones iniciales
[t,x]=ode45(fg,Dt,X0);
   
  
plot(t,x(:,2));
xlabel('Tiempo (Segundos)');
ylabel('Voltaje (Volts)');

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Ameer Hamza 2020-4-29

Can you show the ODEs of your system in mathematical form?

Isaac Husny T 2020-4-29

Modelado del supercapacitor (1).pdf

This is the document with all the ODEs

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Answer 1

David Goodmanson 2020-4-29

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https://ww2.mathworks.cn/matlabcentral/answers/521549-trying-to-solve-time-of-charge-of-supercapacitor-with-ode45#answer_429232

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Hello Isaac

It seems easier to do this as two first order equations for the node voltages V1 and V2 rather than go to a second order differential equation for one of them.

[t V] = ode45(@vdot,[0 1],[0 0])
figure(2)
plot(t,V);grid on
function dVdt = vdot(t,V)
Rc = 1000;
R1 = 100;
R2 = 100;
Rs = Rc + R1;
C1 = 10e-6;
C2 = 100e-6;
E = 5;
i2= (V(1)-V(2))/R2;
i1 = (E - V(1))/Rs - i2;
dVdt = [i1/C1; i2/C2];
end

This calculation does show a time constant on the order of .1 sec.

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Isaac Husny T 2020-4-30

I am required to use a second order equation as shown in the original post, I would appreciate if you can use the same method.

David Goodmanson 2020-5-1

在 MATLAB Online 中打开

Hi Isaac,

I understand that in course work you sometimes required to use certain methods for learning purposes. However, in this case since a much superior method is available, I'm not keen to work out the result for a second order ODE in one variable. Especially since it's complicated and only gives you one voltage, whereas the pair of first order ODEs as used above gives you both. But if you need to do it, here is an outline where the R and C constants are not carried along explicitly (keeping track of those is the hard part). I'm guessing the voltage across C2 is the one of interest.

[1] from the second comment, plug the top two eqns into the bottom two to obtain**
V2dot ~ V1,V2              (a)
V1dot + V2dot ~ E,V1       (b)
[2] differentiate (a) to obtain
V2dotdot ~ V2dot,V1dot     (c)
[3] eliminate V1 from (a) and (b) to obtain
V1dot ~ V2dot,V2,E         (d)
[4] plug (d) into (c) to obtain
V2dotdot ~ V2dot,V2,E      (e), a second order ODE for V2
** in these equations, the right hand side is a linear combination
   of the variables shown.

With standard ode solvers using sets of first order ODEs it seems quixotic to have to go to a second order ODE, but if my grade depended on it I suppose I would do it too.

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